Problem: Evaluate the improper integral if it exists. $\int_{-\infty}^{-1}\dfrac{3}{x^2}\,dx $ Choose 1 answer: Choose 1 answer: (Choice A) A $-3$ (Choice B) B $1$ (Choice C) C $3$ (Choice D) D The improper integral diverges.
Answer: First, let's rewrite the improper integral: $\int_{-\infty}^{-1}\dfrac{3}{x^2}\,dx=\lim_{a\to-\infty}\int_a^{-1} \dfrac{3}{x^2}\,dx$ We can now evaluate the integral: $\begin{aligned} \phantom{\int_1^{\infty}\dfrac1{x^3}\,dx}&=\lim_{a\to-\infty}\int_a^{-1} \dfrac{3}{x^2}\,dx\\ \\ \\ &=3 \lim_{a\to-\infty}\int_a^{-1} \dfrac{1}{x^2}\,dx\\ \\ \\ &=3 \lim_{a\to-\infty}\left[-\dfrac1{x}\right]_a^{-1}\\ \\ \\ &=3\lim_{a\to-\infty}\left(1+\dfrac1{a}\right)\\ \\ &=3\left(\lim_{a\to-\infty}1+\lim_{a\to-\infty}\dfrac1{a}\right)\\ \\ &=3(1+0)\\ \\ &=3 \end{aligned}$ The answer: $\int_{-\infty}^{-1}\dfrac{3}{x^2}\,dx =3$